Simplify and expand the following expression: $ \dfrac{3}{3t - 30}+ \dfrac{5}{3t + 6}+ \dfrac{t}{t^2 - 8t - 20} $
Answer: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $3$ out of denominator in the first term: $ \dfrac{3}{3t - 30} = \dfrac{3}{3(t - 10)}$ We can factor a $3$ out of denominator in the second term: $ \dfrac{5}{3t + 6} = \dfrac{5}{3(t + 2)}$ We can factor the quadratic in the third term: $ \dfrac{t}{t^2 - 8t - 20} = \dfrac{t}{(t - 10)(t + 2)}$ Now we have: $ \dfrac{3}{3(t - 10)}+ \dfrac{5}{3(t + 2)}+ \dfrac{t}{(t - 10)(t + 2)} $ The least common multiple of the denominators is: $ 9(t - 10)(t + 2)$ In order to get the first term over $9(t - 10)(t + 2)$ , multiply by $\dfrac{3(t + 2)}{3(t + 2)}$ $ \dfrac{3}{3(t - 10)} \times \dfrac{3(t + 2)}{3(t + 2)} = \dfrac{9(t + 2)}{9(t - 10)(t + 2)} $ In order to get the second term over $9(t - 10)(t + 2)$ , multiply by $\dfrac{3(t - 10)}{3(t - 10)}$ $ \dfrac{5}{3(t + 2)} \times \dfrac{3(t - 10)}{3(t - 10)} = \dfrac{15(t - 10)}{9(t - 10)(t + 2)} $ In order to get the third term over $9(t - 10)(t + 2)$ , multiply by $\dfrac{9}{9}$ $ \dfrac{t}{(t - 10)(t + 2)} \times \dfrac{9}{9} = \dfrac{9t}{9(t - 10)(t + 2)} $ Now we have: $ \dfrac{9(t + 2)}{9(t - 10)(t + 2)} + \dfrac{15(t - 10)}{9(t - 10)(t + 2)} + \dfrac{9t}{9(t - 10)(t + 2)} $ $ = \dfrac{ 9(t + 2) + 15(t - 10) + 9t} {9(t - 10)(t + 2)} $ Expand: $ = \dfrac{9t + 18 + 15t - 150 + 9t}{9t^2 - 72t - 180} $ $ = \dfrac{33t - 132}{9t^2 - 72t - 180}$ Simplify: $ = \dfrac{11t - 44}{3t^2 - 24t - 60}$